= Solution
For generators $g,h$ of $P$, write $x=g-1$, $y=h-1$. Then
$$
kP\cong k[x,y]/(x^p,y^p).
$$
For each $\lambda\in k$, define a two-dimensional <module> $M_\lambda$ on basis $v,w$ by $xv=w$, $xw=0$, $yv=\lambda w$, $yw=0$. Every product of two of these operators is zero, so the defining relations hold even when $p=2$.
The action of $x$ is one nonzero nilpotent Jordan block. A decomposition of $M_\lambda$ would decompose its restriction to $k[x]/(x^p)$, which is already indecomposable. Thus $M_\lambda$ is indecomposable. An isomorphism $F:M_\lambda\to M_\mu$ commutes with $x$ and $y$, and consequently
$$
\lambda Fx=Fy=yF=\mu xF=\mu Fx.
$$
Since $F$ is invertible and $x\ne0$, $\lambda=\mu$. The infinite <field> supplies infinitely many parameters, proving \b[$kP$ has infinite representation type]. This is the <parameter family for an elementary abelian p-group>.
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